MATLAB: A small problem with interpolation

extrapolation

Dear all,
I have
   A={ [NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]
    [4.8455]    [0.8076]    [0.0864]    [ 0.1917]    [ 0.5184]    [ 0.4275]
    [5.1224]    [0.8537]    [0.3679]    [ 0.7887]    [ 2.2073]    [ 1.8883]
    [5.1932]    [0.8656]    [0.6473]    [ 1.3217]    [ 3.8835]    [ 3.3626]
    [5.2023]    [0.8671]    [0.8252]    [ 1.6275]    [ 4.9510]    [ 4.2929]
    [5.1816]    [0.8636]    [0.7828]    [ 1.6663]    [ 4.6970]    [ 4.0567]
    [5.1818]    [0.8636]    [0.6551]    [ 1.4768]    [ 3.9308]    [ 3.3937]
    [5.1937]    [0.8656]    [0.6070]    [ 1.4121]    [ 3.6422]    [ 3.1528]
    [5.1665]    [0.8611]    [0.6443]    [ 1.5050]    [ 3.8656]    [ 3.3284]
    [5.1786]    [0.8631]    [0.6824]    [ 1.5289]    [ 4.0944]    [ 3.5341]
    [5.0321]    [0.8386]    [0.6997]    [ 1.5754]    [ 4.1985]    [ 3.5209]
    [4.9381]    [0.8231]    [0.7431]    [ 1.6090]    [ 4.4583]    [ 3.6695]
    [4.9326]    [0.8221]    [0.7708]    [ 1.7676]    [ 4.6245]    [ 3.8018]
    [4.7815]    [0.7970]    [0.8851]    [ 2.0084]    [ 5.3107]    [ 4.0625]
    [3.9136]    [0.6523]    [2.5109]    [ 5.3401]    [15.0656]    [ 9.7846]
    [3.8419]    [0.6403]    [4.3700]    [ 8.2200]    [26.2199]    [16.7885]
    [3.8559]    [0.6426]    [4.3385]    [ 7.9125]    [26.0312]    [16.7235]
    [3.8803]    [0.6467]    [3.9299]    [ 7.9730]    [23.5793]    [15.2499]
    [3.8741]    [0.6457]    [3.8246]    [ 8.2150]    [22.9478]    [14.8167]
    [3.8733]    [0.6456]    [3.9232]    [ 8.2825]    [23.5394]    [15.1959]
    [3.8604]    [0.6434]    [3.9364]    [ 8.4879]    [23.6182]    [15.1961]
    [3.8052]    [0.6342]    [3.9610]    [ 8.4783]    [23.7659]    [15.0720]
    [3.8177]    [0.6363]    [4.0807]    [ 8.3607]    [24.4842]    [15.5765]
    [3.7615]    [0.6269]    [5.4267]    [10.9853]    [32.5598]    [20.4115]
    [3.6900]    [0.6150]    [5.8236]    [12.4549]    [34.9416]    [21.4899]
    [3.7249]    [0.6208]    [6.0688]    [12.3084]    [36.4125]    [22.6080]
    [3.7621]    [0.6270]    [6.4914]    [12.5395]    [38.9481]    [24.4177]
    [3.7485]    [0.6248]    [7.1611]    [12.7942]    [42.9669]    [26.8452]
    [3.7492]    [0.6249]    [7.1063]    [12.8555]    [42.6379]    [26.6462]
    [3.7029]    [0.6171]    [6.6088]    [12.9124]    [39.6529]    [24.4730]
    [3.7172]    [0.6195]    [6.3574]    [12.7681]    [38.1441]    [23.6259]
    [3.8148]    [0.6358]    [6.2090]    [12.7283]    [37.2539]    [23.6851]
    [   NaN]    [   NaN]    [   NaN]    [    NaN]    [    NaN]    [    NaN]}
I want to extrapolate the values in the last row.
I tried
 out3=[];
 outtx=CELL2MAT(A)
 for c = 1:size(outtx,2)
     outtxa=inpaint_nans(outtx(:,c)',2);
     outtxa=outtxa';
end
out3=[out3       outtxa ];
but I get the following error
 ?? Subscript indices must either be real positive integers or logicals.
 Error in ==> inpaint_nans at 233
    fda(n,[n, n-1,n+n])=[-2 1 1];
thanks
PS: is there also a way to find the values in the first 4 rows?

Best Answer

It appears to be a bug in this contribution.
You have two options:
  1. report this to the author on the FEX and wait for HIS answer.
  2. Use interp1() with the 'extrap' option
% From my previous answer to your post
xi  = (1:size(A,1))';
% Index the non NaN
idx = ~isnan(A(:,1));
out = interp1(xi(idx),A(idx,:),xi,'linear','extrap');
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